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The angle of rain with vertical = tan^-1(2/6) = 19.6 degree
So for rain to fall vertically on wind screen it must have an angle of 90-18.4 wrt vertical = 71.6 degree

1 mole of oxygen (22.4 Ltr) requires 4Faraday charge= 4*96500 C
Charge supplied = i*t = 1*(160*60+50)=9650C
So vol of O2 produced = 22.4*9650/(4*96500)=22.4/40 ltr

Pressure on surface is 1 atm
Pressure 100 m below 1~ 1+10 = 11 atm
(10 m water column is approx 1 atm)
So final volume of gas will be 1/11th or 9%

This is very scatered data and i don't think that you can find any function thatcan give you an R^2 of 0.99.
Still if you want you can try www.curvefit.com

what is direction of motion of second particle.
As electrical force is more than gravitational force and it is repulsive for identical particles how can we have force of attraction

For adiabatic process P^1+1//Tis constant

So for any gas T final = T initial*(0.5)^^1+1/

Gamma for mono atomic gas is 5/3 and for diatomic gas is 7/4

As initial temperature of both gases is same so for final temperature of both gases are

T_mono/T_di=(0.5)^8/5/(0.5)^11/7=0.5^(1/35) So T_mono <T_di

So B will have higher temp

For refraction at air plastic interface use n/v-1/u = (n-1)/R
(n is refractive index of plastic)
As incident beam is parallel 1/u = 0.
So v = nR/(n-1) = 1.44R/0.44 = 36/11R
But after 2R it gets another refracting surface.
Here u = 36R/11-2R=14R/11
Using 1/v-n/u = (1-n)/(-R)
We can find v

Calculus is true only when dp/p tends to zero. Here it is 0.2. CAlculus will give right answer only when it is used with proper conditions. Here you can't use this result

If i is angle of incidence and r is angle of reflection and t is thickness of slab than a simple ray diagram can show that the deflection will be given by t*(tan(i)-tan(r))*cos(i)
i is given. Taking n=1.5 find r and solve for deflection

Let refractive index of material be n.
So n = sin((A+da)/2)/sin(A/2)
where A is angle of prism = 60
da is angle of least deviation = 37
This gives n

For the case of water (refractive index = 1.33) refractive index of medium is n/1.33.
So n/1.33 = sin((A+dw)/2)/sin(A/2)
From this solve for dw

Consider a spherical shell of radius R carrying a charge Q and there is no charge in vicinity. What is the potential energy of system.

To find potential energy of such a system we assume taht at any instant the shell contains charge q and find amount of work done (dW) to bring infinitisimal charge dq on the shell. Then we integrate from q=0 to Q. Negative of this will give me potential of system. Now this potential energy is there because of repulsion of charge of same body (the shell) and is not because of interaction between two bodies (like two point charges)
Such potential energy is called self energy

You can check this post and that may make things clear

http://www.goiit.com/posts/list/electricity-self-energy-and-potential-energy-1525.htm#5944

Perfect answer bumba

NO fuel is required. And it may be well within the effect of gravity. Once you give orbital velocity to a body tangentially it will keep on revolving for ever. (Otherwise who is supplying fuel to moon)

Nivedh is right. If a projectile just passes an obstcle of height 'h' then it may be maximum height but that is not necessary

Perfect approach Betu.